What is the sum from n=0 to infinity of 2^{n+1} ?

The sum from n=0 to infinity of 2^{n+1} is diverges

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sum from n=0 to infinity of 2^{n+1}

n = 0 \sum \infty 2^{n + 1}

diverges

Solution steps

n = 0 \sum \infty 2^{n + 1}

Simplify

2^{n + 1} : 2^{n} \cdot 2

= n = 0 \sum \infty 2^{n} \cdot 2

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 0 \sum \infty 2^{n}

Apply Series Geometric Test:

diverges

= 2 diverges

= diverges

n = 0 \sum \infty 4 (\frac{1}{3})^{n}

n = 0 \sum \infty \frac{4 ^{n}}{2 ^{n}}

n = 2 \sum \infty \frac{7}{n ( n + 2 )}

n = 1 \sum \infty \frac{n + 2}{n ^{2}}

n = 1 \sum \infty \frac{1}{( n + 1 ) n}

What is the sum from n=0 to infinity of 2^{n+1} ?
The sum from n=0 to infinity of 2^{n+1} is diverges